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Probability - density

  1. Mar 24, 2010 #1
    1. The probability density function for the continuous random vector X = (X1, X2) is given by:

    Fx1, x2(X1,X2) = { 1/2 if X1+X2[tex]\leq[/tex] 2,X1[tex]\geq[/tex] 0,X2[tex]\geq[/tex]0
    0 otherwise }

    Calculate the probability that X1>2X2



    I do not know which formular to use. I can't find the right one. How do I do this?:confused: The joint density formular?
     
  2. jcsd
  3. Mar 24, 2010 #2

    LCKurtz

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    Draw a picture in the x1x2 plane showing where the density is 1/2. Then, on the same picture, draw the region in question where x1>2x2. That should show you the region to work with.
     
  4. Mar 24, 2010 #3
    Yeah, I did that. However, I don't know which formular to use.
     
  5. Mar 24, 2010 #4

    Dick

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    You integrate the probability density over the region where x1>2*x2.
     
  6. Mar 24, 2010 #5
    So joint density was right?
     
  7. Mar 24, 2010 #6

    Dick

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    If that's what you call integrating probability density over part of your region where x1>2*x2, then sure.
     
  8. Mar 24, 2010 #7
    Joint density would be:

    F x,y(x,y)=xe-x(y+1), x>0 ,y=0
     
  9. Mar 24, 2010 #8

    Dick

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    That's not it. The region where density is 1/2 is a triangle. Does the picture you drew show that?
     
  10. Mar 24, 2010 #9
    Well, I actually didn't know that this one can not be used for triangles?
    There's another formular I found for joint density...

    F N,X(n,x)=Xne-2x/n!
     
  11. Mar 24, 2010 #10

    Dick

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    Stop looking up formulas. You don't need them. It's not that complicated. LCKurtz said to draw a picture. You said you did. What does it look like? Describe the triangle where density is 1/2.
     
  12. Mar 24, 2010 #11

    LCKurtz

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    Huh?? Where did that come from? You gave us the joint density in the original post. And the probability that X1>2X2 should be a number between 0 and 1.
     
  13. Mar 25, 2010 #12
    Well, I eventually solved it myself. Integration wasn't needed.
     
  14. Mar 25, 2010 #13

    Dick

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    Good. That's always the best way to solve it. And, yes, you don't need explicit integration since the probability density isn't a function of x1 and x2. You just find the overlap area and multiply by 1/2.
     
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